Precalculus
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Eliminate the parameter (What does that mean?) and write a rectangular equation for (could it be [t^2 + 3][2t]?) x= t^2 + 3 y = 2t Without a calculator (how can I do that?), determine the exact value of each expression. cos(Sin^1
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I need help with I just can't seem to get anywhere. this is as far as I have got: Solve for b arcsin(b)+ 2arctan(b)=pi arcsin(b)=pi2arctan(b) b=sin(pi2arctan(b)) Sub in Sin difference identity let 2U=(2arctan(b))
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The original question I had was write arcsin4 in the form a+ib. I manage and understand how to get so far BUT How do I get from cosacoshbisinasinhb=4 to 2m(pi)+/ iarccosh4 arcsin4 = a + b i > 4 = sin(a + bi) sin(a + bi) =
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Please can you help me with this question? Choose the option which is a false statement: A arctan(tan2/3pi))=1/3pi B arccos(cos(3/4pi))=3/4pi C sin(arcsin(1/2pi))=1/2pi D arcsin(1/2squareroot3)=1/3pi E arcsin(sin(3/4pi))=1/4pi
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I am given arcsin (sin 4pi) do the arc sin and sin cancel out? How do we solve this?
asked by MONDAY on November 6, 2017 
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Differentiate. y= (cos x)^x u= cos x du= sin x dx ln y = ln(cos x)^x ln y = x ln(cos x) (dy/dx)/(y)= ln(cos x) (dy/dx)= y ln(cos x) = (cos x)^x * (ln cos x) (dx/du)= x(cos x)^(x1) * (sin x) =  x sin(x)cos^(x1)(x)
asked by Chelsea on March 9, 2011 
calculus
Differentiate. y= (cos x)^x u= cos x du= sin x dx ln y = ln(cos x)^x ln y = x ln(cos x) (dy/dx)/(y)= ln(cos x) (dy/dx)= y ln(cos x) = (cos x)^x * (ln cos x) (dx/du)= x(cos x)^(x1) * (sin x) =  x sin(x)cos^(x1)(x)
asked by Chelsea on March 10, 2011 
Calculus
s=çdx/(4+5cos x). By using tsubstitution, i.e. t=tan(x/2) we get cos x=(1t^2)/(1+t^2) and dx=2dt/(1+t^2). Substituting in s and simplifying, we get s= 2çdt/4(1+t^2)+5(1t^2)=2çdt/(9t^2). Using standard result
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Find the exact value of sin(A+B) when angle A is arcsin(8/17) and angle B is arcsin(5/13)
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